1. tìm x
a.\(\dfrac{1}{5}+x=\dfrac{13}{30}\)
b.\(\dfrac{1}{6}-x=\dfrac{5}{12}\)
c. \(x.\left(\dfrac{-1^{ }}{2}\right)^2=\dfrac{1}{16}\)
d. \(x:\dfrac{7}{11}=\dfrac{9}{33}\)
e. \(\dfrac{3}{5}.x=\dfrac{-21}{10}\)
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a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)
b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)
c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)
d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)
b, \(x+\dfrac{2}{3}\) = \(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\)
\(x+\dfrac{2}{3}\) = \(\dfrac{23}{30}\)
\(x\) = \(\dfrac{23}{30}\) - \(\dfrac{2}{3}\)
\(x\) = \(\dfrac{1}{10}\)
a, \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x\) = \(\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x\) = \(\dfrac{1}{4}\)
x = \(\dfrac{2}{3}-\dfrac{1}{4}\)
\(x=\dfrac{5}{12}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
1)
a.\(\dfrac{1}{5}+x=\dfrac{13}{50}\)
\(\Leftrightarrow x=\dfrac{13}{50}-\dfrac{1}{5}=\dfrac{13-10}{50}=\dfrac{3}{50}\)
b.\(\dfrac{1}{6}-x=\dfrac{5}{12}\)
\(\Leftrightarrow x=\dfrac{1}{6}-\dfrac{5}{12}=\dfrac{2-5}{12}=-\dfrac{3}{12}=-\dfrac{1}{4}\)
c.\(x\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow x\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}.\left(-\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d.\(x:\dfrac{7}{11}=\dfrac{9}{33}\)
\(\Leftrightarrow x=\dfrac{9}{33}.\dfrac{7}{11}=\dfrac{3}{11}.\dfrac{7}{11}=\dfrac{21}{121}\)
e.\(\dfrac{3}{5}.x=-\dfrac{21}{10}\)
\(\Leftrightarrow x=-\dfrac{21}{10}:\dfrac{3}{5}=-\dfrac{21}{10}.\dfrac{5}{3}=-\dfrac{7}{2}\)